Question: Let $f(x) = -6x^{2}-3x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-6x^{2}-3x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -6, b = -3, c = 1$ $ x = \dfrac{+ 3 \pm \sqrt{(-3)^{2} - 4 \cdot -6 \cdot 1}}{2 \cdot -6}$ $ x = \dfrac{3 \pm \sqrt{33}}{-12}$ $ x = \dfrac{3 \pm \sqrt{33}}{-12}$ $x =\dfrac{3 \pm \sqrt{33}}{-12}$